Chapter 33: Electromagnetic Waves
33-27
THINK Electromagnetic waves travel at the speed of light, and carry both linear
momentum and energy.
EXPRESS The speed of the electromagnetic wave is c  f , where  is the wavelength
and f is the frequency of the wave. The angular frequency is   2f , and the angular
wave number is k  2 /  . The magnetic field amplitude is related to the electric field
amplitude by Bm  Em / c. The intensity of the wave is given by Eq. 33-26:
I
1
c 0
2
Erms

1
Em2 ,
2c0
where Erms  Em / 2.
ANALYZE (a) With  = 3.0 m, the frequency of the wave is
f 
c 2.998  108 m / s

 10
.  108 Hz.

3.0 m
(b) From the value of f obtained in (a), we find the angular frequency to be
  2f  2   Hz)  6.3  108 rad / s.
(c) The corresponding angular wave number is k 
2
2

 2.1 rad / m.
  m
(d) With Em = 300 V/m, the magnetic field amplitude is
Bm 
Em
300V/m

 1.0 106 T.
c
2.998 108 m/s


(e) Since E is in the positive
y
direction,
must be in the positive z direction so that
B

their cross product E  B points in the positive x direction (the direction of propagation).
(f) The intensity of the wave is
Em2
(300V/m)2
I

 119W/m2  1.2 102 W/m 2 .

8
20c 2(4  H/m)(2.998 10 m/s)
(g) Since the sheet is perfectly absorbing, the rate per unit area with which momentum is
delivered to it is I/c, so
dp IA (119 W / m2 )(2.0 m2 )


 8.0  107 N.
8
dt
c
2.998  10 m / s
dp / dt 8.0  107 N
(h) The radiation pressure is pr 

 4.0  107 Pa.
2
A
2.0 m
LEARN The energy density in this case is given by
u
I
119 W/m 2

 4.0 107 J/m3
c 2.998 108 m/s
which is the same as the radiation pressure pr.
33-37
THINK A polarizing sheet can change the direction of polarization of the incident beam
since it allows only the component that is parallel to its polarization direction to pass.
EXPRESS The 90° rotation of the polarization direction cannot be done with a single
sheet. If a sheet is placed with its polarizing direction at an angle of 90° to the direction
of polarization of the incident radiation, no radiation is transmitted.
ANALYZE (a) The 90° rotation of the polarization direction can be done with two
sheets. We place the first sheet with its polarizing direction at some angle , between 0°
and 90°, to the direction of polarization of the incident radiation. Place the second sheet
with its polarizing direction at 90° to the polarization direction of the incident radiation.
The transmitted radiation is then polarized at 90° to the incident polarization direction.
The intensity is
I  I 0 cos 2  cos 2 (90   )  I 0 cos 2  sin 2  ,
where I 0 is the incident radiation. If  is not 0° or 90°, the transmitted intensity is not
zero.
(b) Consider n sheets, with the polarizing direction of the first sheet making an angle of 
= 90°/n relative to the direction of polarization of the incident radiation. The polarizing
direction of each successive sheet is rotated 90°/n in the same sense from the polarizing
direction of the previous sheet. The transmitted radiation is polarized, with its direction of
polarization making an angle of 90° with the direction of polarization of the incident
radiation. The intensity is
I  I 0 cos 2 n (90 / n) .
We want the smallest integer value of n for which this is greater than 0.60I0. We start
with n = 2 and calculate cos2 n (90 / n) . If the result is greater than 0.60, we have obtained
the solution. If it is less, increase n by 1 and try again. We repeat this process, increasing
n by 1 each time, until we have a value for which cos2 n (90 / n) is greater than 0.60. The
first one will be n = 5.
LEARN The intensities associated with n = 1 to 5 are:
I n 1  I 0 cos 2 (90)  0
I n  2  I 0 cos 4 (45)  I 0 / 4  0.25 I 0
I n 3  I 0 cos 6 (30)  0.422 I 0
I n  4  I 0 cos8 (22.5)  0.531I 0
I n 5  I 0 cos10 (18)  0.605 I 0 .
Thus, I > 0.60I0 with 5 sheets.
33-53
THINK The angle with which the light beam emerges from the triangular prism depends
on the index of refraction of the prism.
EXPRESS Consider diagram (a) shown next. The incident angle is  and the angle of
refraction is 2. Since 2    90 and   2  180, we have
 2  90    90 
1

180     .
2
2
(a)
(b)
ANALYZE Next, examine diagram (b) and consider the triangle formed by the two
normals and the ray in the interior. One can show that  is given by
  2(  2 ) .
Upon substituting /2 for 2, we obtain   2(   / 2) which yields   (  ) / 2.
Thus, using the law of refraction, we find the index of refraction of the prism to be
n
sin  sin 12 (  )

.
sin  2
sin 12 
LEARN The angle  is called the deviation angle. Physically, it represents the total
angle through which the beam has turned while passing through the prism. This angle is
minimum when the beam passes through the prism “symmetrically,” as it does in this
case. Knowing the value of  and  allows us to determine the value of n for the prism
material.